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What are the roots of the equation x^2-6x+58=0 in simplest a+bi form?

Sagot :

irspow

Answer:

Step-by-step explanation:

[tex]x^2-6x+58=0\\ \\ \text{Since the discriminant, 36-232<0, there are no real roots, only two imaginary ones.}\\ \\ x=3\pm 7i\\ \\ (x+3+7i)(x+3-7i)[/tex]

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